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How come, for any n > 2, the nth triangular number + the nth square

  1. Nov 5, 2013 #1
    number cannot be prime? I have checked this for n from 3 to 53,509, the latter being the limit for unsigned int. I believe this is true, and I thereby claim that this is a true statement. However, I don't see any obvious explanation for it.
     
    Last edited: Nov 5, 2013
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  3. Nov 5, 2013 #2

    Mentallic

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    The nth triangle number Tn is

    [tex]T_n = \frac{n(n+1)}{2}[/tex]

    and the nth square number Sn is

    [tex]S_n = n^2[/tex]

    And so
    [tex]T_n+S_n = \frac{n(n+1)}{2}+n^2=\frac{n(3n+1)}{2}[/tex]

    Can you now show why this expression must be composite (not prime) for all n?
     
  4. Nov 5, 2013 #3
    Well, it wouldn't be true for all n. Only for n > 2, assuming my statement is correct. :biggrin:
     
  5. Nov 5, 2013 #4
    When n is odd and more than 2, 1 / 2 * (3n + 1) is a whole number bigger than 1, and so the result is composite. When n is even and more than 2, n * (3n + 1) is even, so (n / 2) * (3n + 1) is even. Therefore the sum is composite for all n > 2 is correct. :biggrin: Many thanks for the help.
     
    Last edited: Nov 5, 2013
  6. Nov 5, 2013 #5

    Mentallic

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    Well, ignoring the fact that you included the criteria that n>2 in your proof below, [itex]n\leq 2[/itex] would also spit out composite numbers, right? Your proof only considers that n is even or odd which means that all integers [itex]n\leq 2[/itex] would also be involved. The only reason to restrict yourself to n>2 is such that we have a meaningful square and triangle number.

    Adding to the end of that: because we then have a product of two integers, mainly n and [itex]\frac{3n+1}{2}[/itex].

    This is incorrect. If [itex]n(3n+1)[/itex] is even, then [itex]\frac{n}{2}(3n+1)[/itex] isn't necessarily even, but rather an integer. But most importantly, you haven't proven that the expression is a product of two integers and hence composite.


    Also, while it's not absolutely necessary, when you consider n to be even, you could let n=2k for some integer k and substitute that into your expression, then show that the result is composite, and similarly for n odd, let n=2k+1.
     
  7. Nov 5, 2013 #6
    Many thanks. :wink:
     
    Last edited: Nov 5, 2013
  8. Nov 6, 2013 #7
    You do get a prime number for n=1 or n=2, and the proof also uses the fact that n>2, when it says
    so I don't see what the problem is here.
     
  9. Nov 6, 2013 #8

    Mentallic

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    Haha yeah I thought about it while out today and realized the criteria n>2 is necessary, which goldust even incorporated into his proof! Sorry about that goldust.
     
  10. Nov 6, 2013 #9
    Many thanks for the help! :biggrin: The proof is a bit trickier than I initially thought. :eek: When n is even and more than 2, n / 2 is an integer more than 1, and 3n + 1 is also an integer more than 1, so n / 2 * (3n + 1) ends up being divisible by both n / 2 and 3n + 1. Cheers! :biggrin:
     
    Last edited: Nov 6, 2013
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